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Let $A, B\subset\mathbb{R}$ two Lebesgue-measurable sets of positive Lebesgue-measure. Prove that $A-B$ contains an interval.

I thought that being the measure positive then the Hausdorff dimension of the two sets is 1. But I don't know how to go further. The hint says: "use the convolution of the two characteristics function of the two sets" but I can't really figure out how to use it.

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It is enough to consider the case when $A$ and $B$ are bounded. Let $f=\chi_A * \chi_B$. It is a general fact that convolution of two functions in $L^{1} \cap L^{\infty}$ is continuous. [This is proved by approximating these functions by continuous with compact support in $L^{1}$ norm]. Now $\int f(x)dx=\int \chi_A(x)dx \int \chi_B(x)dx=m(A)m(B) >0$. Hence $f(x_0) >0$ for some $x_0$ and there exists $\epsilon >0$ such that $f(x) >0$ for all $x \in (x_0-\epsilon ,x_0+ \epsilon)$. So for any $x$ in this interval there exists $y$ such that $\chi_A(x-y)\chi_B(y) >0$ which means $y \in B \cap (x-A)$. It follows that $x =(x-y)+y \in A+B$.

We have proved that $A+B$ contains an interval. For $A-B$ just change $B$ to $-B$.

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